# QM interpretation of Larmor precession

Typically when I think of a spin moment in an external field, I visualize it classically in terms of a magnetic moment precessing around the external field vector.

Now how is this described quantum mechanically? My guess would be that the external field causes transitions between the up-state and down-state at the Larmor frequency. However if the Zeeman energy was much greater than the thermal energy, wouldn't the spin need to stay in the low energy state? Does that mean the spin ceases precession in the classical picture?

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Larmor precession is quantum mechanically described by the interaction of the spinor with the magnetic field. Without magnetic field the spin-up eigenstates states in the x and y direction are given by:

$S_x^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ 1\end{array}\right) \exp(-iEt)$

$S_y^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ i\end{array}\right)\exp(-iEt)$

The direction of the spin in the xy-plane is given by the angle between the two spinor components in the complex plane.

The two components get an extra energy factor from the magnetic field B but with a different sign:

$S^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}\exp(-iE_Bt) \\ \exp(+iE_Bt)\end{array}\right)\exp(-iEt)$

The result is that the spin direction rotates (precesses) in time.

Regards, Hans

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